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Let $Q_{\alpha}(X)$ for $\alpha \in (0,1)$ be the $\alpha$-th quantile of the random variable $X$. Suppose that $X$ and another random variable $Y$ (independent of $X$) have nice continuous (or maybe even smooth) densities. Can I get an upper bound on the difference: $$ \left| Q_{\alpha}(X+Y)- Q_{\alpha}(X) \right| $$ in terms of the densities of $X$ and $Y$?

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Basic Approach. Note that we can write

$$ Q_\alpha(X) = F_X^{-1}(\alpha) $$

where $F_X(x) \equiv P(X < x)$ is the cumulative distribution function (CDF) of $X$. So the expression in the absolute value is

$$ F_{X+Y}^{-1}(\alpha)-F_X^{-1}(\alpha) $$

where the corresponding PDF

$$ f_{X+Y} = f_X \ast f_Y $$

That is, $f_{X+Y}$ is the convolution of $f_X$ and $f_Y$. (Thanks to A.S. in the comments for catching this, and shame on me for not noticing for over four years!)

Does that help? Or do you need something more specific?

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  • $\begingroup$ But $F_{X+Y}\neq F_X*F_Y$. pdfs, not cdfs, convolute. $\endgroup$ – A.S. Oct 10 '15 at 5:41
  • $\begingroup$ @A.S.: I'm astounded that I didn't see this comment (or the mistake that precipitated it). Thanks for the catch, and I'll edit. $\endgroup$ – Brian Tung Jan 25 '20 at 0:09

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